Answer the following : The pH of rainwater collected in a certain region of Maharashtra on a particular day was 5.1. Calculate the H⊕ ion concentration of the rainwater and its percent dissociation. - Chemistry

The pH of rainwater collected in a certain region of Maharashtra on a particular day was 5.1. Calculate the H+ ion concentration of the rainwater and its percent dissociation.

Solution

Given: pH of rainwater = 5.1

To find:

i. H+ ion concentration

ii. Percent dissociation

Formula:

i. pH = -log10[H3O+]

ii. Percent dissociation =  α × 100

Calculation: From the formula (i),

pH = -log10[H3O+]

∴ log10[H3O+] = -5.1

= -5 - 0.1 + 1 - 1

= (-5 - 1) + 1 - 0.1

= -6 + 0.9 = bar(6).9

∴ [H3O+] = Antilog10[bar(6).9]

= 7.943 × 10-6 M

Considering that the pH of rainwater is due to the dissociation of a monobasic strong acid (HA), we have

$\ce{HA_{(aq)} + H2O_{(l)} -> H3O^+_{ (aq)} + A^-_{ (aq)}}$

∴ [H3O+] = α

α = 7.943 xx 10^-6

From formula (ii),

Percent dissociation = 7.943 xx 10^-6 xx 100 = 7.943 xx 10^-4

i. H+ ion concentration is 7.943 xx 10^-6 M

ii. Percent dissociation is 7.943 xx 10^-4.

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APPEARS IN

Balbharati Chemistry 12th Standard HSC Maharashtra State Board
Chapter 3 Ionic Equilibria
Exercise | Q 4.08 | Page 62